题目链接:HDU - 1512

Once in a forest, there lived N aggressive monkeys. At the beginning, they each does things in its own way and none of them knows each other. But monkeys can't avoid quarrelling, and it only happens between two monkeys who does not know each other. And when it happens, both the two monkeys will invite the strongest friend of them, and duel. Of course, after the duel, the two monkeys and all of there friends knows each other, and the quarrel above will no longer happens between these monkeys even if they have ever conflicted.
Assume that every money has a strongness value, which will be reduced to only half of the original after a duel(that is, 10 will be reduced to 5 and 5 will be reduced to 2).
And we also assume that every monkey knows himself. That is, when he is the strongest one in all of his friends, he himself will go to duel.
Input
There are several test cases, and each case consists of two parts.
First part: The first line contains an integer N(N<=100,000), which indicates the number of monkeys. And then N lines follows. There is one number on each line, indicating the strongness value of ith monkey(<=32768).
Second part: The first line contains an integer M(M<=100,000), which indicates there are M conflicts happened. And then M lines follows, each line of which contains two integers x and y, indicating that there is a conflict between the Xth monkey and Yth.
Output
For each of the conflict, output -1 if the two monkeys know each other, otherwise output the strongness value of the strongest monkey in all friends of them after the duel.
 
题意描述:有n只猴子,每只猴子有一个值,两只猴子如果打架的话,他们的值各自掉一半。给出m个事件,每个事件给出两只猴子,如果两只猴子不认识的话,打完架就成为了朋友(两只猴子的各自朋友也都互相成为了朋友),求出打完架后两只猴子的所有朋友中值最大的。
算法分析:这道题我刚开始做的时候,想到了并查集,以为这就够了(其实时间复杂度我也不敢直视),交上去果断TLE了,后来看了讨论里有说,尼玛,这就是传说中的左偏树的题目啊,果断得好好学习一下,弥补一下自己的数据结构的知识。
说明:头一次搞左偏树,代码是借鉴别人的,不过真心写的比较好就拿来了。同时,推荐一下集训队的论文,基本上看了论文后就对左偏树有了一定的了解了。

左偏树的特点及其应用

 /*左偏树*/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#define inf 0x7fffffff
using namespace std;
const int maxn = +; int father[maxn];
struct node
{
int l,r;
int dis;
int strong;
}LTree[maxn];
int Find(int x)
{
if (father[x]==x) return x;
return father[x]=Find(father[x]);
}
int Merge(int x,int y)
{ //返回合并后的根
if (x==) return y;
if (y==) return x;
if (LTree[x].strong < LTree[y].strong) //大顶堆
swap(x,y);
LTree[x].r = Merge(LTree[x].r,y); //递归合并右子树和Y
int l = LTree[x].l , r = LTree[x].r;
father[r] = x; //更新T右子树的根
if (LTree[l].dis < LTree[r].dis) //维护堆性质
swap(LTree[x].l,LTree[x].r);
if (LTree[x].r == ) //如果没有右子树 则距离为0
LTree[x].dis = ;
else
LTree[x].dis = LTree[LTree[x].r].dis + ;
return x;
}
int del(int x)
{ //返回删除根以后左右子树的合并的根
int l,r;
l=LTree[x].l;
r=LTree[x].r;
father[l]=l;
father[r]=r;
LTree[x].l=LTree[x].r=LTree[x].dis=;
return Merge(l,r);
}
void solve(int x,int y)
{
LTree[x].strong /= ;
LTree[y].strong /= ;
//问每次PK以后,当前这个群体里力量最大的猴子的力量是多少。
int left,right;
left = del(x);
right = del(y);
left = Merge(left,x);
right = Merge(right,y);
left = Merge(left,right);
printf("%d\n",LTree[left].strong);
}
int main()
{
int n,m,x,y;
while (scanf("%d",&n)!=EOF)
{
for (int i= ;i<=n ;i++)
{
scanf("%d",&LTree[i].strong);
LTree[i].l=;
LTree[i].r=;
LTree[i].dis=;
father[i]=i; //起始已自己为父亲
}
scanf("%d",&m);
for (int i= ;i<=m ;i++)
{
scanf("%d%d",&x,&y);
int fx=Find(x),fy=Find(y);
if (fx == fy) printf("-1\n");
else solve(fx,fy);
}
}
return ;
}

hdu 1512 Monkey King 左偏树的更多相关文章

  1. hdu 1512 Monkey King —— 左偏树

    题目:http://acm.hdu.edu.cn/showproblem.php?pid=1512 很简单的左偏树: 但突然对 rt 的关系感到混乱,改了半天才弄对: 注意是多组数据! #includ ...

  2. HDU 1512 Monkey King (左偏树+并查集)

    题意:在一个森林里住着N(N<=10000)只猴子.在一开始,他们是互不认识的.但是随着时间的推移,猴子们少不了争斗,但那只会发生在互不认识 (认识具有传递性)的两只猴子之间.争斗时,两只猴子都 ...

  3. HDU 1512 Monkey King ——左偏树

    [题目分析] 也是堆+并查集. 比起BZOJ 1455 来说,只是合并的方式麻烦了一点. WA了一天才看到是多组数据. 盲人OI (- ̄▽ ̄)- Best OI. 代码自带大常数,比启发式合并都慢 [ ...

  4. HDU 1512 Monkey King(左偏堆)

    爱争吵的猴子 ★★☆ 输入文件:monkeyk.in 输出文件:monkeyk.out 简单对比 时间限制:1 s 内存限制:128 MB [问题描述] 在一个森林里,住着N只好斗的猴子.开始,他们各 ...

  5. ZOJ2334 Monkey King 左偏树

    ZOJ2334 用左偏树实现优先队列最大的好处就是两个队列合并可以在Logn时间内完成 用来维护优先队列森林非常好用. 左偏树代码的核心也是两棵树的合并! 代码有些细节需要注意. #include&l ...

  6. zoj 2334 Monkey King/左偏树+并查集

    原题链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1389 大致题意:N只相互不认识的猴子(每只猴子有一个战斗力值) 两只 ...

  7. HDU1512 ZOJ2334 Monkey King 左偏树

    欢迎访问~原文出处——博客园-zhouzhendong 去博客园看该题解 题目传送门 - ZOJ2334 题目传送门 - HDU1512 题意概括 在一个森林里住着N(N<=10000)只猴子. ...

  8. hdu1512 Monkey King(左偏树 + 并查集)

    Once in a forest, there lived N aggressive monkeys. At the beginning, they each does things in its o ...

  9. LuoguP1456 Monkey King (左偏树)

    struct LeftTree{ int l,r,val,dis; }t[N]; int fa[N]; inline int Find(int x){ return x == fa[x] ? x : ...

随机推荐

  1. 孤荷凌寒自学python第六十四天学习mongoDB的基本操作并进行简单封装3

    孤荷凌寒自学python第六十四天学习mongoDB的基本操作并进行简单封装3 (完整学习过程屏幕记录视频地址在文末) 今天是学习mongoDB数据库的第十天. 今天继续学习mongoDB的简单操作, ...

  2. CodeForces-757B Bash's Big Day

    题目链接 https://vjudge.net/problem/CodeForces-757B 题目 Description Bash has set out on a journey to beco ...

  3. HDU 4027 Can you answer these queries(线段树 + 观察 )

    这题主要考察观察能力. 2^63最多只需要开7次根号就会变成1,当数字变成1之后就不需要再对其进行操作. 对于含有大于1数字的区间,向下更新. 对于数字全为1的区间,直接返回. #include &l ...

  4. 以太坊源码分析(52)以太坊fast sync算法

    this PR aggregates a lot of small modifications to core, trie, eth and other packages to collectivel ...

  5. 中英文混截,一个中文相当于n个英文

    项目中遇到这么个需求,截取中英文字符串,一个中文相当于2个英文,全英文时截取12个英文字母,全中文时是6个中文汉字,中英文混合时是12个字节,在网上有找到这样的解决方案,但我没能静下心来研究懂,于是自 ...

  6. python 读取数据库时,datetime类型无法被json序列化--解决方案

    新增针对datetime的jsonencode: # -*- coding: utf-8 -*- import json from datetime import date, datetime cla ...

  7. 模拟Windows系统“回收站”

    HTML: <!DOCTYPE html><html> <head> <meta http-equiv="content-type" co ...

  8. 解决jquery与zepto等其它库冲突兼容的问题

    解决jquery与zepto等其它库冲突兼容的问题;(function ($) {    }) (jQuery); ;(function ($) {    }) (Zepto); 在Bootstrap ...

  9. ARM嵌入式开发中的GCC内联汇编__asm__

    在针对ARM体系结构的编程中,一般很难直接使用C语言产生操作协处理器的相关代码,因此使用汇编语言来实现就成为了唯一的选择.但如果完全通过汇编代码实现,又会过于复杂.难以调试.因此,C语言内嵌汇编的方式 ...

  10. win7 iis 7.0 碰到 503错误,找到的解决方案

    Service Unavailable HTTP Error 503. The service is unavailable. 今天要布署一个网站,在自己的电脑上,结果碰到服务器503错误,找应用程序 ...